如图所示,锐角三角形ABC的内心为I,过点A作直线BI的垂线,垂足为H,点E为圆I与边CA的切点.

(1)见解析 (2) 25°


解:(1)由圆I与AC相切于点E得IE⊥AC,结合HI⊥AH,得∠AEI=∠AHI=90°,所以A,I,H,E四点共圆.
(2)由(1)知A,I,H,E四点共圆,所以∠IEH=∠HAI.由题意知∠HIA=∠ABI+∠BAI= ∠ABC+ ∠BAC= (∠ABC+∠BAC)= (180°-∠C)=90°- ∠C,结合IH⊥AH,得∠HAI=90°-∠HIA=90°-(90°- ∠C)= ∠C,所以∠IEH= ∠C.由∠C=50°得∠IEH=25°.